A uniform current carrying ring of mass \(m\) and radius \(R\) is connected by a massless string as shown. A uniform Magnetic field \(B_0\) exist in the region to keep the ring in horizontal position, then the current in the ring is

\((a)\) \(\frac{mg}{\pi R B_0}\)

\((b)\) \(\frac{mg}{ R B_0}\)

\((c)\) \(\frac{mg}{3 \pi R B_0}\)

\((d)\) \(\frac{mg}{\pi R^2 B_0}\)

No vote yet

8 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe tension of the string is \(T\), and the weight force on the loop is \(mg\). Because of the symmetry of the loop, the force caused by the magnetic field on the left half of the loop is equal and opposite to the force caused by the magnetic field on the right half. So the two net forces acting on the loop are \(T\) and \(mg\). The loop is at rest, so they must be equal.

\(T=mg\).

The torque on the loop due to the string is \(RT\), and the torque on the loop due to the magnetic field is \(B_0 A I\), where \(A\) is the area of the loop and \(I\) is the current. The area of the loop is \(A=\pi R^2\), so the torque due to the magnetic field is \(\pi R^2 B_0 I\). The loop is at rest, so the two torques must be equal.

\(RT=\pi R^2 B_0 I\)

\(Rmg=\pi R^2 B_0 I\)

\(I=\frac{mg}{\pi R B_0} \quad (a) \).

Log in to reply

T=mg (Force Balance).

TR=i\(\pi\)\(R^{2}\)\(B_{0}\) (Torque Balance)

Solve the two to get answer as (a).

Log in to reply

Is ur Age 14 Mr. Advitiya B. ?

Log in to reply

Comment deleted Jul 25, 2014

Log in to reply

Did u Qualify in NTSE ?

Log in to reply

Comment deleted Jul 25, 2014

Log in to reply

I am curious ...becoz... My age person .in my Country.. . went to high Levels in challenges in Less TIME ... U joined from March 2013 right

Log in to reply

c) T + (2

PIRBI) = Mg ....... BALANCING THE FORCES TR=(PIRRBi) ..................BALANCING TORQUE SOLVE TO GET i :)Log in to reply